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glowball
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My 90 percent solution to the Rusduck challenge is further below.

Background:
The Fitch Cheney trick (spectator picks any five cards and magician sets one aside and magician arranges other 4 cards which are taken to his assistant in another room who divines the set aside card) is brilliant but has one major weakness which is the magician picks the target card.

Rusduck put out a challenge for magicians to come up with a way for 4 randomly freely picked cards to be arranged face up so that his assistant could look at those four cards and determine a fifth randomly freely chosen card that was set aside out of view. No external signals allowed, just the four cards used as a signaling means. The four cards given as a packet to the assistant.

Note I (glowball) added that last sentence but I believe it's in the spirit of his challenge.

The mathematically difficult part of this challenge is:
1. randomly freely picked 4 cards to be face up
2. the spectator freely picks the fifth card (the target card, emphasis on the spectator picking the target not the magician)

Over the years numerous magicians have come up with solutions that are very ingenious but don't entirely meet Rusduck's criteria. Some of these solutions involve turning up to three of the signal cards face down on the table.

I would not be surprised if someone else has come up with this idea too long before me.

My 90 percent solution:
All normal modern playing card decks have 22 one way faces:
Excluding the diamonds the other three suits each have seven one way cards:
Ace, three, five, six, seven, eight, nine
therefore 7 times 3 suits equals 21 one way cards. The diamond suit has only one which is the seven of diamonds so we add that giving us 22 one way cards.

The odds of one signal card being a one-way card is 22/52 is about 42 percent which is not good odds but there are four signal cards chosen so this greatly increases the odds of getting at least one one-way card amongst the four. I'm going to stick my finger in the air and say 90% of the time the magician will have a one-way card amongst the four and in situations where there is none the assistant can equivocate between two cards "your card is one of two cards it's either the five of clubs or the jack of clubs, it isn't the five is it" regardless of their answer the assistant says "I thought so".

The specifics of my solution are (using SHoCkeD):
The magician arranges the four cards as follows:
1. holds the four cards in a fan facing himself
2. temporarily moves the highest card to the far right of the fan so he can easily focus on the other three cards.
3. uses the other three cards to indicate a value of 1-6 ie LMH is 1, LHM is 2, MLH is 3 etc. Note that if the spectator set aside target card is greater than six then the magician mentally subtracts six from that target card to arrive at the 1-6 value. If the set aside target card is a king just treated as though it's a queen.
4. to indicate the suit of the target card: now position the highest card (that was temporarily moved to the far right of the four card fan) using the Stewart James principle to insert that card amongst the other three cards in the appropriate SHCD position.
5. starting from the left look for the first one-way card and make sure the middle pip is pointing down if the target card is ace through six and make sure the pip is pointing up if the target card is a seven through king.

That's it, but this packet must be consistently held vertically the correct way by the spectator when delivering it to the magician's assistant (not an issue if the four cards are face up on the table in a row but I do not like the table method because there are 100 possible external signals that the assistant could use and an astute audience knows this).

What the assistant does after receiving the packet:
1. makes sure that the packet is correctly held vertically so that it is not upside down.
2. looks for the highest card (uses SHCD for any tie breaking) and determines the suit based on the relative position of that card and mentally repeats the suit name a few times to cement the suit name and then jogs that card down halfway to get it out of view while he evaluates the other three cards.
3. evaluates the other three cards LMH etc to come up with the 1-6 value.
4. next starting from the left of the four cards he looks for the first one-way card (including the jogged down suit indicator card) and then if the middle pip is pointing up he adds six to the 1-6 value. If there is no one way card then the assistant does the equivocate between the two possible cards.
5. If the value is 12 then the assistant does an equivocate between the queen and the king.

The worst case scenario is that the 1-6 value is 6 and there is no one way card amongst the four cards. In this rare situation the target card that the spectator picked could be the king, or queen, or six thus two equivocate would be needed. Therefore we have an additional rule that if the target is a king and there is no one way card then the magician does an additional final step: he turns the last card horizontally so it's face to face with the rest of the packet (I know that's totally cheating but hey maybe it's better than having to do two equivocates).
With this rule in place the assistant will know the one equivocate to use.
Another solution to this problem is just to remove the kings from the deck ahead of time which also eliminates the need to ever equivocate between the queen and king.

Well my solution certainly has some weaknesses but 80 to 90 percent of the time it meets the Rusduck criteria.

I do have another Rusduck criteria solution in progress that works 100 percent of the time (it has been tested), it has the following features:
1. uses Sam's club red-backed bicycle cards that are unaltered (neither the faces nor the backs nor the sides).
2. All 52 cards are unique.
3. the target card is freely chosen by the spectator
4. the four signal cards are freely chosen by the spectator
5. there are no external signals
6. there are no forces nor equivocates
7. the unaltered faces of the four cards tell everything

Stay tuned.
glowball
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There is another solution to those nasty kings: instead of removing the 4 kings from the deck leave them in and remove the 4 sevens and then change the methodology that the magician and his assistant use ie instead of subtracting six from the high cards to determine the 1-6 number if the target card is greater than a seven then subtract seven to get the 1-6 number. Then the magician's assistant if they see the middle pip pointing up (on the first one way card) they add seven to the 1-6 number.

The reason this is better is that people will never notice that the sevens are missing from the deck whereas they very well might notice that the kings are missing.
glowball
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I didn't know how to calculate the true odds of getting at least one one-way card amongst the four cards so I called my brother, David Finley, who is a retired college professor (PhD in mathematics) and asked what the answer was and after him saying something about a Poisson distribution and something else about not using binomial nomenclature he said to use a hyper geometric method none of which I understood (I got a B in algebra and geometry but never even took calculus). He said that one hit odds 22/52 calculation was fine for one card but to calculate for four cards the simpler way was to calculate the odds of not getting a hit with four cards and then subtract from 1. Therefore:
(30/52)*(29/51)*(28/50)*(27/49)=0.10122818358
which is almost exactly 10 percent. Then just subtract that result from 1.00000 giving 0.89877181641 which is surprisingly close to my initial guess of 90 percent!
glowball
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If using a borrowed deck a way to get rid of the 4 sevens would be to openly remove the four sevens and use them as 4 detectives (the signal cards). The spectator does pick one card (the target card). See my thread about using the four sevens:

https://www.themagiccafe.com/forums/view......forum=99

Then after doing the trick one time with the four sevens the magician just openly puts the 4 sevens in his pocket saying to the spectator "what if you pick the four detectives" and then the magician uses this 90 percent solution method (spectator picks the four signal cards and picks the target card). The magician repeats this trick two or three times.

With the sevens out of the way an equivocate is only needed 10 percent of the time (when there is no one-way card amongst the four signal cards).
Claudio
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There are many ways to calculate the odds of drawing at least one one-way card amongst a set of any 4. Glowball’s approach is the simplest but here’s a different one which relies on 4-card combinations and might be of interest to some as being more intuitive (calculations presented below assume, rightly, that the order of the 4 cards is unimportant).

a. How many combinations of 4 cards are there in a 52-card deck? 50*49*48*47/4*3*2 = 270 725
b. How many combinations of 4 cards only belonging to the 30-card subset are there? 30*29*28*27/4*3*2 = 27 405
c. How many combinations of 4 cards with at least one one-way card are there? 270 725 – 27 405 = 243 320
d. What are the odds of drawing any set of 4 cards with at least one-way card? 243320/270725 = 0.898772

The odds of drawing one set of 4 cards with at least one-way card is 0.898772 which is about 90% as established previously.

Interesting approach and very good effort, but as it does not meet the 100% success rate it’s not quite useable as an effect it its current state. However, if you don’t mind disregarding the purist mathematical approach then it’s guaranteed 100%. For instance, use the original Cheney’s method if there’s no one-way card. The assistant will know that’s the method to be used because there’s no one-way card in the set. The magician would have to choose the card themselves, of course (though there are ways to force it). Many other subtle handling can be used to reach the desired outcome.
glowball
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I believe you can reliably do this method in public because the magician's assistant has the equivocate as an "out". Just one equivocate between two cards is still pretty impressive. However as you and others have said (and I agree) this type of trick is not a great magician's trick. And I would add not great for the general public therefore I would never do this trick in a major show.

See Manjul and Cindy do the Fitch Cheney trick very successfully with a great reaction before a math oriented audience for the Momath Museum using supergiant playing cards (go to the 7:00 minute mark):

https://youtu.be/oGEcYkwF6d4

The five cards were picked by a clean spectator.
Of course the trick is greatly heightened if the audience thinks the five cards were shuffled and positioned by a second random person.

The academic community loves this type of trick especially if math is a major part of the solution.

Back to the 90% Rusduck discussion. I do like your suggestion (better than using my equivocate "out") to use the Fitch Cheney method as an "out" if there is no one-way card amongst the four signal cards however there is a catch 22 that needs to be resolved. At the start of the trick we are hoping to get at least one one-way card but do we have the spectator first
a. randomly pick five cards and then pick the target card from the five? or
b. do we have the spectator pick four signal cards and then pick a target card?

Either way we have a problem of morphing into the Fitch Cheney trick. If we do "method a" then we would need two one-way cards amongst the five because of the possibility that the spectator could pick the one-way card as the target card (then we would not have it available to use in the 4 signal cards) and we would need to morph into Fitch Cheney but then it would be too late to morph into Fitch Cheney because the target card was already selected by the spectator.

If we use "method b" and we have the spectator first pick the four cards and then the magician sees there is no one-way card, now how do we add the fifth card and reselecting from amongst the five cards which four cards are the signal cards without it looking weird? after the spectator picks the four cards does the magician say "pick one more card because we need five cards" then the magician can pick the target card from amongst the five?

Reminder to all that we are only doing the Fitch Cheney trick here as an "out" because we did not get any one-way card and note that the Fitch Cheney method requires the magician to pick the target card from amongst the five cards selected by the spectator.

I think the magician should have pre-instructions to the spectator before they start selecting any cards otherwise it may appear the magician is changing the rules on the fly. Ideally the pre-instructions should be something like "in a little bit you will select four cards to be our detectives and then select another card to be the secret card". However if there is no one-way card amongst the four cards now the instructions have to be changed to let the magician pick the secret card from amongst the five cards.

Hmmm, maybe we can be a little vague with the pre-instructions and still the instructions can sound specific "in a little bit we're going to play a detective game where you select five cards and some of them will be detectives". With this kind of pre-instruction it was not stated who would pick the secret card and not stated which cards would be the detectives. That statement should work when a one-way card is amongst the four cards and I think that same statement would work for doing the Fitch Cheney trick.

Next the magician must not say "pick the four detectives" instead says the generic statement "pick any 4 cards from anywhere in the deck" then if the magician sees that one of the four cards is a one-way card says "now pick another card from the deck to be the secret card".

If there is no one-way card amongst the four cards the magician says "we need one more card so pick another card from the deck". Now the magician can select the target card via Fitch Cheney (the better way is to say "these four cards will be our detectives" and the magician removes those four cards to be taken to the assistant so by default the remaining fifth card is the secret target card). The statement "we need one more card" is a little awkward but should suffice because it is NOT saying that this one more card is the target, this statement is just saying we need an additional card.
glowball
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And since the magician had earlier said "where you will pick five cards" then it does not seem unusual to say pick a fifth card but it is a little strange because why didn't the magician just say "pick five cards" in the first place.
Claudio
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Unfair to use a stooge for a math demo Smile

I don’t see any problem here. For instance, you could say:

“We’re going to play a game of Cluedo with very simple rules. To start with we need 5 cards. Please pick any five cards from this shuffled deck.”

The cards are then shown face up and you have to adapt your patter and put your magician hat on to deal with the different scenarios.

The good news with picking 5 cards is that the odds of getting at least one one-way card among them rise to nearly 95%.

a) If there is only one-way card, say “Now, we have to choose the victim and the other cards will be clues for my assistant who’ll play the detective”.
It should not be too difficult to force any card but the one-way card to portray the victim.

b) If there are two or more one-way cards, use the same line as above and the spec. has a free choice.

c) If there are no one-way card, you say something like:
“OK, you’ve chosen 5 cards, 4 will be clues and this one for example (you pick the correct suit) will be the murder victim. My assistant will play the detective”
Or, you can force any of the duplicate (suit) cards.
glowball
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My response to a):

"a) If there is only one-way card, say 'Now, we have to choose the victim and the other cards will be clues for my assistant who will play the detective'.
It should not be too difficult to force any card but the one-way card to portray the victim.'"

The way I see using the Fitch Cheney "method a)" as an "out" for my Rusduck 90 percent trick:

Method a) pros: raises the probability of getting at least one one-way card to over 99.67 percent of the time, method b) and c) can still be used where possible, when needed the Fitch Cheney method is nice (even though the spectator did not specify the target card)

Method a) cons: the vast majority of time "method a)" will have to be used every time when performing the trick and even though

"It should not be too difficult to force any card but the one-way card to portray the victim"

this takes away from the purity of the selection of the target card (will happen most of the time).

Now to contrast the above "out" method with using an equivocate for an "out" (should there be no one-way card) pros and cons:

equivocate pros: 90 percent of the time the spectator's selections are clean and pure and the trick is performed the same way, the equivocate is easy to do when needed

equivocate cons:
10 percent of the time the audience will like the effect less than if the assistant nailed it (the question is how much less), the four Kings or the four Sevens need to be removed from the deck ahead of time

To magicians out there do you agree with my evaluation of the pros and cons?

Or do you think that Claudio and I are just having fun brainstorming? or both?
glowball
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Oops, it's 94.51 percent not 99.67 (I used the wrong formula). Claudio's "about 95 percent" statement is correct.
glowball
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Another pro for Claudio's "method a)" is that by having the spectator pick all five cards at once (instead of four then one as I have suggested) is that you not only increase the odds of one one-way card but you also increase the odds of at least two one-way cards. And if you get two one-way cards your presentation is very pure and you can then say "set one of those cards aside as the jewel thief" ie the target card.
Claudio
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Here are the odds:

Nearly 95% of the time there’ll be at least one one-way card, but 71% of the time* there’ll be at least two or more one-way cards among a set of 5, therefore no forcing will be required.

I can’t see any problem with the forcing when there’s only one-way card. Place it at the bottom of the packet and fan the five cards and ask the spec. to touch a card. The chances are that the one-way card will not be picked and therefore whatever card has been picked will be considered the “victim”. If the one-way card were to be picked, say something like : “This is our first clue, you need to pick 3 more.”

* feed the formula below (in bold) into https://web2.0calc.com/ to see the formula expanded and the result.
1 - ((ncr(30,5))/ncr(52,5)) - ( (ncr(22,1)*ncr(30,4))/ncr(52,5) = 71%
glowball
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Thanks Claudio for that 71% calculation, this is good news, I needed that!

Another way to utilize "method a)" without the magician ever touching the 5 cards during the selection process:

Magician says "we're going to play a game of detectives. My assistant has left the room so please pick any five cards and have them face up on the table".

If the magician sees one and only one one-way card he then says:
"please move one card to the side"
If that one card happens to be the one-way card then the magician says:

"this will be Columbo now pick three more detectives to complete his team and the final fifth card will be the jewel thief".

If the first card that was moved to the side is not the one-way card the magician then says:

"this will be the Jewel thief the remaining four cards are detectives".

Note that the "Columbo" wording will only be needed 20 percent of the time for "method a)".
Claudio
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I reread Glowball’s first post and in particular this paragraph:

Quote:
90% of the time the magician will have a one-way card amongst the four and in situations where there is none the assistant can equivocate between two cards "your card is one of two cards it's either the five of clubs or the jack of clubs, it isn't the five is it" regardless of their answer the assistant says "I thought so".


It dawned on me that instead of equivocating, the performer could simply take out of the deck the 2nd card of the ambiguous pair and add it to the spectator’s selected card as an added challenge.

Obviously, all this is done openly so that the audience understands it (and some script is necessary) but without the assistant being aware of this. However, as soon as the assistant studies the cards, she'll recognize the situation and unprompted will announce both cards.

Looks good on paper as it seems to turn a weakness into strength and it's always good have another string to one's bow.
hcs
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I’m sure some people here know that Card Shark's Phoenix deck is a one-way deck on both sides.
glowball
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To hcs: I did not know that, thanks much. I have several Phoenix decks.
To all:
I had prepared the below post before I saw hcs's post about the Phoenix decks:
With a deck like the below (all the cards have one-way faces) I can use my combined one-way faces method 100% of the time, no need for any "outs" and no need to remove the kings or the sevens ie Rusduck criteria 100% pure solved:

Korean Joseon Kingdom Folk Painting Art Premium Standard Size Unique Playing Poker Cards, Deck of Cards for Kids & Adults
$19.99
Amazon.com - Seller

Also there are many decks that have one-way face cards such as Star Trek and Alice in wonderland etc. these are not 100% Rusduck but close 99.67 percent of getting at least one one-way card. I wonder what percent of the time that the magician will get two or more one-way cards when the deck now has 12 more one-way cards than just 22 ie has 34 one-way cards. I also wonder what the percent is of getting at least three one-way cards amongst the five? The advantage of having three one-way cards is that one is for protection of being wasted by being picked as the target. The last of the other two one-way cards can be used as a reference by the first one-way card so the magician doesn't have to worry about the spectator inadvertently turning the whole packet upside down.

Alice in wonderland deck of cards where the face cards are one-way cards:

https://www.etsy.com/listing/534730762/v......land-old

Also my post several days ago in this thread above my "stay tuned" comment are statements based on the fact that some Sam's Club bicycle decks have one way faces on all the cards. I will start another more detailed thread about that entitled "Rusduck Challenge 100% solved".
glowball
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Please see my 100% solution thread (no outs needed, no equivocates needed)

https://www.themagiccafe.com/forums/view......forum=99
glowball
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See the following 2 threads about one-way facers which are also called pointer cards. All the posts appear to be about the magician performing by himself to find a selected card that has been reversed or similar such effects.

I didn't see any posts that used the one-way principle to encode a selected card similar to the Fitch Cheney trick. However there are some interesting posts in these two threads such as the fact that older bicycle spade face cards (JS, QS, KS) had a hard to detect one-way design.

https://themagiccafe.com/forums/viewtopi......orum=201

https://themagiccafe.com/forums/viewtopi......orum=37a
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